CAT 2019 Slot 1 Question Paper

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

The sixth performance was composed by Badal. Hence C is the answer.

Since the dancers performed their second items in the same sequence of their performance of their first items. The table will be as follows:

The items assigned by Ashman were performed consecutively. The number of performances between items assigned by each of the remaining composers was the same.

Also, the first items performed by the four dancers were all assigned by different music composers. Badal can come only at the place as shown in the table.

Then Ashman can only compose for the following performances.

Hence Dyu will compose for the following performances:

From (i) No composer who assigned item to Princess, assigned any item to Queen.
From (ii) No composer who assigned item to Rani, assigned any item to Samragni.

Hence Dyu will compose for Samragni 1st Performance and Gagan will compose for Queen 1st Performance. Also, Badal will compose for Samragni 2nd Performance and Dyu will compose for Queens 2nd Performance.

Hence, the complete table is as follows:

The first and the sixth items were composed by Badal. Hence D is the answer.

It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the item b should be at least twice. The minimum number of items will be obtained when a=1 and b=99, which means there are only two different types of items.

It is given that the most expensive item is a diamond ring of type a and there is exactly one of these. Since the number of items of type b should be at least twice of that of a and the number of items of type c should be at least twice of that of b and so on. So the maximum number of different types of items of a, b and c will be obtained when a=1, b=2, c=4, d=8, e=16, f=69. Hence the maximum number of different types of items will be 6.

If the number of items is 7, then the minimum number of prizes should be 1+2+4+8+16+32+64=127 which is more than 100.

Hence 6 is the answer.

Option A: There are exactly 75 items of type e.

a=1,b=2,c=4,d=8, e=85. Here the maximum value of e= 85. Hence it can take the value 75.

An example of such case is a=1,b=2,c=4,d=18, e=75

Option B: There are exactly 30 items of type b.

a=1 b=30 and c=69. Hence this case is also possible.

Option C: There are exactly 45 items of type c.

Since the value of d should be at least 90, it means that d is not present because 45+90 will be more than 100(maximum number of items). Only a,b and c are present.

The maximum value of b = 22 and a =1, but 45+22+1=68, which is less than 100. So this case is not possible.

Option D: There are exactly 60 items of type d.

d=60, c=30, b=9 and a=1. a+b+c+d=100. Hence this case is possible.

C is the answer.

The total number of items from 1 to 100, which are of same type as in box 45 = 31+1+43=75

Now to maximize the number of items, a=1, b=2, c=4, d=18 and e=75(given)

There can be maximum 5 types of items.

If we consider number of items to be 6, then minimum number of items of 5th type will be 16, 1+2+4+8+16+75=106 which is more than 100.

Let the speed of cars be a and b and the distance =d

Minimum time taken by 1st car = 6 hours,

For maximum difference in time taken by both of them, car 1 has to start at 10:00 AM and car 2 has to start at 11:00 AM. 

Hence, car 2 will take 5 hours. 

Hence a= $$\ \frac{\ d}{6}$$ and b = $$\ \frac{\ d}{5}$$

Hence the speed of car 2 will exceed the speed of car 1 by $$\ \dfrac{\ \ \frac{\ d}{5}-\ \frac{\ d}{6}}{\ \frac{\ d}{6}}\times\ 100$$ = $$\ \dfrac{\ \ \frac{\ d}{30}}{\ \frac{\ d}{6}}\times\ 100$$ = 20

We have, $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$$

Now, $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}}$$ = $$\frac{\sqrt{a_2} - \sqrt{a_1}}{(\sqrt{a_2} + \sqrt{a_1})(\sqrt{a_2} - \sqrt{a_1})}$$   (Multiplying numerator and denominator by $$\sqrt{a_2} - \sqrt{a_1}$$)

= $$\frac{\sqrt{a_2} - \sqrt{a_1}}{({a_2} - {a_1}}$$

=$$\frac{\sqrt{a_2} - \sqrt{a_1}}{d}$$   (where d is the common difference)

Similarly, $$ \frac{1}{\sqrt{a_2} + \sqrt{a_3}}$$ = $$\frac{\sqrt{a_3} - \sqrt{a_2}}{d}$$ and so on.

Then the expression $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$$

can be written as $$\ \frac{\ 1}{d}(\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_3}+..........................\sqrt{a_{n+1}} - \sqrt{a_{n}}$$

= $$\ \frac{\ n}{nd}(\sqrt{a_{n+1}}-\sqrt{a_1})$$ (Multiplying both numerator and denominator by n)

= $$\ \frac{n(\sqrt{a_{n+1}}-\sqrt{a_1})}{{a_{n+1}} - {a_1}}$$     $$(a_{n+1} - {a_1} =nd)$$

= $$\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$$

Since AB is a diameter,  AQB and APB will right angles.

In right triangle APB, AP = $$\sqrt{10^2-6^2}=8$$

Now, 2AQ=AP  => AQ= 8/2=4

In right triangle AQB, AP = $$\sqrt{10^2-4^2}=9.165$$ =9.1 (Approx)

We have, $$(5.55)^x = (0.555)^y = 1000$$

Taking log in base 10 on both sides,

x($$\log_{10}555$$-2) = y($$\log_{10}555$$-3) = 3

Then, x($$\log_{10}555$$-2) = 3.....(1)

y($$\log_{10}555$$-3) = 3 .....(2)

From (1) and (2)

=> $$\log_{10}555$$=$$\ \frac{\ 3}{x}$$+2=$$\ \frac{\ 3}{y}+3$$

=> $$\frac{1}{x} - \frac{1}{y}$$ = $$\frac{1}{3}$$